Red and Black

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 23 Accepted Submission(s) : 13

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can‘t move on red tiles, he can move only on black tiles.<br><br>Write a program to count the number of black tiles which he can reach by repeating the moves described above. <br>

Input

The input consists of multiple data sets. A data set
starts with a line containing two positive integers W and H; W and H are the
numbers of tiles in the x- and y- directions, respectively. W and H are not more
than 20.<br><br>There are H more lines in the data set, each of
which includes W characters. Each character represents the color of a tile as
follows.<br><br>‘.‘ - a black tile <br>‘#‘ - a red tile
<br>‘@‘ - a man on a black tile(appears exactly once in a data set)
<br>

Output

For each data set, your program should output a line
which contains the number of tiles he can reach from the initial tile (including
itself). <br>

Sample Input

6 9

....#.

.....#

......

......

......

......

......

#@...#

.#..#.

11 9

#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#[email protected]#.#.

.#.#####.#.

.#.......#.

.#########.

...........

11 6

..#..#..#..

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

..#..#..#..

7 7

..#.#..

..#.#..

###.###

[email protected]

###.###

..#.#..

..#.#..

0 0

Sample Output

45

59

6

13

简单题意：

　　#为障碍，"."为陆地，@表示人，求这个人最大能走到少范围

思路分析：

　　简单搜索问题，，，最值得注意的是，先输入的是列， 而不是行

# include <iostream>
# include <queue>
# include <cstring>
# include <fstream>
using namespace std; // 好坑呀， 先输入列 在输入行

struct Info
{
    int x;
    int y;
}start;
int n, m;
char map[101][101];
int is[101][101];
int dir[4][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}};

int dfs();
bool border(Info & info)
{
    if(info.x >= 1 && info.x <= m && info.y >= 1 && info.y <= n)
    return true;
    return false;
}
int main()
{
    //fstream cin("aaa.txt");
    while(cin >> n >> m)
    {
        if(n == 0 && m == 0)
        break;
        memset(is, 0, sizeof(is));
        for(int i = 1 ; i <= m; i++)
            for(int j = 1; j <= n; j++)
            {
                cin >> map[i][j];
                if(map[i][j] == ‘@‘)
                {
                    start.x = i;
                    start.y = j;

                }
            }
        is[start.x][start.y] = 1;
        map[start.x][start.y] = ‘.‘;
        cout << dfs() << endl;
    }
    return 0;
}
int dfs()
{
    int jishu = 1;
    queue <Info> Q;
    Q.push(start);
    Info now, next;
    while(!Q.empty())
    {
        now = Q.front();
        Q.pop();

        for(int i = 0; i < 4; i++)
        {
            next.x = now.x + dir[i][0];
            next.y = now.y + dir[i][1];
            if(!border(next))
                continue;
            if(map[next.x][next.y] == ‘#‘)
            continue;
            if(is[next.x][next.y])
            continue;

            jishu++;
            is[next.x][next.y] = 1;
            Q.push(next);
        }
    }
    return jishu;
}